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General Information
Power Factor Correction Questions and Answers
 Back to Basics: What does Power Factor Mean and Why Must We
Correct it?
Power factor is the ratio between the kW and the kVA drawn by an
electrical load where the kW is the actual load power and the kVA is the
apparent load power.
Simply, it is a measure of how efficiently the load current is being
converted into useful work output and more particularly is a good indicator
of the effect of the load current on the efficiency of the supply system.
Consider this:
When you buy fuel for a vehicle, the manufacturer makes in it in litres, the
pump dispenses it in litres and you pay for it in litres. Dollars/litre – simple!
When you buy potatoes, the supplier bags them in kilos the shop sells them
in kilos and you pay for them in kilos. Dollars/kg – simple!
When you buy electricity, the “manufacturer” (electricity generator) makes
kVA (kilo volt amperes) and you pay for it in kWh (kilowatt hours) or maybe
on your bill (Units) – not so simple!
Maybe we all should have kVA meters to make life simple.
So what is the kilowatt hour (or unit) we get on our bills? Simply, 1000
watts of electricity being used for 1 hour.
Example: 10 x 100 watt lamps x 1 hour=1000 watts/hr divided by
1000=1kWh – simple!
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Now here comes the problem: In an alternating current (AC) electrical
supply, a mysterious thing called “Power Factor” comes into play. Power
Factor is simply the measure of the efficiency of the power being used, so,
a power factor of 1 would mean 100% of the supply is being used efficiently.
A power factor of 0.5 means the use of the power is very inefficient or
wasteful.
So what causes Power Factor to change? In the real world of industry and
commerce, a power factor of 1 is not obtainable because equipment such as
electric motors, welding sets, fluorescent and high bay lighting create what
is called an “inductive load” which in turn causes the amps in the supply to
lag the volts. The resulting lag is called Power Factor.
For a 3 phase power supply: kVA, which the electricity generator
makes=Line Volts x Amps x 1.73÷ 1000. This is converted to kilowatts kW by
the formula: Line Volts x Amps x 1.73 ÷ 1000 x Power Factor=kW (V x A x
1.73÷ 1000 x pf) or kVA x pf=kW (N.B. 1.73 is the square root of 3) so as
the power factor worsens from say 0.98 to 0.5, the generator has to supply
more kVA for each kW you are using.
For example, a large electric motor will typically have a Power Factor of
about 0.85 at full load. If we have a hypothetical electric motor rated at
100kW, then ignoring the inherent inefficiency of the motor, when running
at full load the electricity supplier would have to supply 100÷0.85=118kVA to
provide the 100kW to run the motor.
Or put the other way they would be supplying 18% more electricity than
they are charging you for. If the same motor was operating “off load” at say
50kW or being used on a cyclic duty then the power factor may go as low as
0.5.
In this case the supplier would have to supply “double” the kVA to match
the 50kW duty point. (50÷0.5=100kVA)
How this power is wasted can be shown graphically since in 3 phase power
supplies "power" can be represented and measured as a triangle. ACTIVE
Power is the base line and is the “real” usable power measured and paid for
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in kW. REACTIVE power is the vertical or that part of the supply which
causes the inductive load. The reactive power in is measured in kVAr (kilo
volt-amperes reactive). APPARENT Power is the hypotenuse. This is the
component the electricity generator must supply and it is the resultant of
the other two components, measured in kVA. Mathematically the power can
be calculated by pythagoras or trigonometry whereby Power Factor is
expressed as COS phi Ø (The angle between Apparent Power and Active
power)
But we want a simple explanation so consider a barge being pulled by a
horse:
Since the horse cannot walk on water its pulling effort is reduced by the
“angle” of the tow rope.
If the horse could walk on water then the angle Phi Ø would be zero and
COSINE Ø=1. Meaning all the horse power is being used to pull the load.
However the relative position of the horse influences the power. As the
horse gets closer to the barge, angle Ø1 increases and power is wasted, but,
as the horse is positioned further away, then angle Ø2 gets closer to zero
and less power is wasted
So, by improving Power Factor (reducing the angle), the reactive power
component is reduced
What does it do to my electricity bill?
As stated above in a 3 phase power supply, kW consumed is 3 phase VOLTS
x AMPS x 1.73 x Power Factor. The Electricity Company supply you VOLTS x
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AMPS and they have to supply extra to make up for the loss caused by poor
Power Factor.
When the power factor falls below a set figure, the electricity supply
companies charge a premium on the kW being consumed, or, charge for the
whole supply as kVA by adding reactive power charges (kVar) to the bill.
 What are the Causes of Poor Power Factor?
Today’s commercial, industrial, retail and even domestic premises are
increasingly populated by electronic devices such as PCs, monitors, servers
and photocopiers which are usually powered by switched mode power
supplies (SMPS). If not properly designed, these can present non-linear
loads which impose harmonic currents and possibly voltages onto the mains
power network.
Harmonics can damage cabling and equipment within this network, as well as
other equipment connected to it. Problems include overheating and fire risk,
high voltages and circulating currents, equipment malfunctions and
component failures, and other possible consequences.
A non-linear load is liable to generate these harmonics if it has a poor power
factor. Other loads can present poor power factors without creating
harmonics. This post looks at these issues, the circumstances that can lead
to damaging harmonic generation, and practical approaches to reducing it.
 The two causes of poor power factor
At the simplest level, we could say that an electrical or electronic device’s
power factor is the ratio of the power that it draws from the mains supply
and the power that it actually consumes. An ‘ideal’ device has a power factor
of 1.0 and consumes all the power that it draws. It would present a load that
is linear and entirely resistive: that is, one that remains constant
irrespective of input voltage, and has no significant inductance or
capacitance. Fig. 1. Shows the input waveforms that such a device would
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exhibit. Firstly, the current waveform is in phase with the voltage, and
secondly both waveforms are sinusoidal.
Fig 1: Input voltage and current waveforms for a device with PF = 1.0
In practice, some devices do have unity power factors, but many others do
not. A device has a poor power factor for one of two reasons; either it
draws current out of phase with the supply voltage, or it draws current in a
non-sinusoidal waveform. The out of phase case, known as ‘displacement’
power factor, is typically associated with electric motors inside industrial
equipment, while the non-sinusoidal case, known as ‘distortion’ power factor,
is typically seen with electronic devices such as PCs, copiers and battery
chargers driven by switched-mode power supplies (SMPSs). We shall look
briefly at the displacement power factor before moving on to the distortion
case, which is of more immediate concern to electronic power system
designers. However it is important to be aware of both cases. For example,
some engineering courses discuss the power factor issue only in terms of
motors, which causes confusion when their students later encounter poor
power factor as exhibited by an SMPS.
 Electric motors and displacement power factor problems
Electric motors create powerful magnetic fields which produce a voltage, or
back emf, in opposition to the applied voltage. This causes the supply
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current to lag the applied voltage. The resulting out of phase current
component cannot deliver usable power, yet it adds to the facility’s required
supply capacity and electricity costs. Fitting capacitors across motors
reduces the phase lag and improves their power factor.
 SMPSs and distortion power factor problems
Fig 2: Non-sinusoidal current waveform drawn by SMPS with poor power
factor
Whereas displacement power factor loads do not cause harmonics and their
associated problems, distortion power factor loads such as SMPSs will do so
unless their power factor is improved.
An SMPS’s AC front end typically comprises a bridge rectifier followed by a
large filter capacitor. This circuit only draws current from the mains when
the line voltage exceeds that across the capacitor. This causes current to
flow discontinuously, resulting in the non-sinusoidal current waveform shown
in Fig. 2.
It is possible to use Fourier transforms, a mathematical process, to analyse
this waveform and break it down into a set of sinusoidal components. These
comprise the fundamental frequency – 50 Hz in Europe, 60 Hz in America –
and a set of predominantly odd multiples of the fundamental, known as
harmonics. The third harmonic is 150 (or 180) Hz, the fifth, 250 (300) Hz
and so on. Fig. 3 shows a typical harmonic spectrum for an electronic SMPS
load. The fundamental component is usefully consumed by the SMPS, while
the harmonics are reactive and create the problems described above. The
ratio of the fundamental amplitude to the sum of all the harmonic
amplitudes gives the device’s power factor.
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Fig 3: Typical harmonic spectrum for an electronic SMPS load
 International standard
An international standard exists to describe and set acceptable limits for a
product’s mains harmonics generation. Within the EU, its reference is IEC
61000-3-2, covering equipment power levels from 75 W to 600 W. The
standard assigns equipment into four Classes – A, B, C and D. Class D covers
personal computers, personal computer monitors and television receivers.
 How does Power Factor Correction work?
Considering most of the load with the users results in Inductive (lagging )
Power Factor, by installing suitably sized switched capacitors into the power
distribution circuit, the Power Factor is improved and the value becomes
nearer to 1 thus minimising wasted energy, improving the efficiency of a
plant,
 Who gets the savings?
First of all we must understand that savings happen due to reduction in
Current drawn hence maximum savings occur due reduction in line Losses in
the transmission of power.
This means maximum savings are accrued by the Authorities involved in
Transmission and distribution of energy to the user.
Depending on the supply Authority they will either charge the user to cover
their losses in the Transmission lines by charging in KVA as in NSW or
impose penalties for Energy used at poor Power Factor.
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Also All users in Australia pay for installation capacity in terms of Peak
Demand. Peak Demand is measured in KVA. Hence a Poor power factor will
result in User requiring more KVA for the same amount of energy used.
Alternatively an improvement in power Factor will result in liberating more
kW from the available supply and save the user money!
 What is the Pay Back Period of Investment Made?
The purchase cost of the installation is usually repaid in 1 to 4years in
electricity savings. This depends on how the Electricity suppliers charge for
the energy supplied and how optimum is the use of the Energy by the user.
It is important therefore to make measurements to collect accurate
information regarding the current Power Factor and install appropriately
sized Power Factor correction equipment.
Also for large and Multiple users an Improved Power Factor enables the
user to go back with the improvements made and negotiate a better Energy
rate from the Energy suppliers.
 Why is there a need of putting Harmonic and line filters in Power
Factor Correction Equipment?
Distortion in Supply due to Harmonics can result in Higher current being
drawn by Capacitors. Line filters ensure that this effect is minimized
thereby prolonging the life of the installation.
In addition some of the supply authorities use Power distribution lines also
for metering purposes. This signal for measurement is done at higher
frequencies and will be disturbed by the Power Factor Correction capacitors
if suitable filters are not installed. Most supply authorities now demand
that such filters are included with the Power Factor Correction Equipment.
 What should be the rating of capacitors used in Power Factor
Correction equipment?
During switching of capacitors and due to Fluctuation of Voltages in the
supply Power Factor correction capacitors are subjected to very high
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voltages. It is therefore important to select Power Factor correction
equipment which can sustain these conditions. It is prudent to select
capacitors which can withstand 480V 3Phase AC supply in normal cases and
in sites prone to high voltage fluctuations 525V 3 Phase rating is
appropriate.
(This document is under development and if you have any questions please
feel free to raise it with R.P Switchboards for inclusion in the document.)
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